# Arithmetic Mean – Definition, Formula, Properties & Examples

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In statistics, the Arithmetic Mean (AM) or average is one of the representative figures along with the median and mode. These figures, i.e., arithmetic mean, median, and mode are also called measures of central tendencies. The arithmetic mean is the ratio of the sum of all observations to the total number of observations.

Let’s understand what is arithmetic mean and how it is calculated along with its properties using examples.

## What is The Arithmetic Mean?

Arithmetic mean which is also called an arithmetic average or in short the mean or the average. It is calculated by adding all the values in a given data set and then dividing it by the total number of items within that set. The arithmetic mean is calculated using various methods, which are based on the amount and the distribution of the data.

In a physical sense, the arithmetic mean can be thought of as the centre of gravity. For example, you want to visit a place during a particular period of time and want to know the temperature of a place during that period. In such cases, we find the average or mean temperature for days during that period for the past few years.

Let’s for example, if we collect the temperatures of the past 10 days, which were found to be 25, 21, 23, 25, 24, 26, 21, 20, 24, and 22 (in degrees Celsius), then the mean temperature of these 10 days is $\frac{25 + 21 + 23 + 25 + 24 + 26 + 21 + 20 + 24 + 22}{10} = \frac{231}{10} = 23.1^{\circ}\text{C}$.

## Properties of Arithmetic Mean

Let us have a look at some of the important properties of the arithmetic mean.

Suppose we have $n$ observations represented by $x_{1}$, $x_{2}$, $x_{3}$, …, $x_{n}$ and $\overline{x}$ is their arithmetic mean, then

• If all the observations in the given data set have a value say $m$, then their arithmetic mean is also $m$. For example, if we have $5$ observations as $12$, $12$, $12$, $12$ and $12$. So, their mean will be $\frac{12 + 12 + 12 + 12 + 12}{5} = \frac {60}{5} = 12$.
• The algebraic sum of deviations of a set of observations from their arithmetic mean is zero, i.e., $\left ( x_{1} − \overline{x} \right) + \left ( x_{2} − \overline{x} \right) + \left ( x_{3} − \overline{x} \right) + … + \left ( x_{n} − \overline{x} \right)= 0$. For discrete data, $\sum \left(x_{i} – \overline{x} \right) = 0$ and for grouped frequency distribution, $\sum f\left(x_{i} – \overline{x} \right) = 0$.
• If each value in the data increases or decreases by a fixed value, then the mean also increases/decreases by the same number. If the mean of $x_{1}$, $x_{2}$, $x_{3}$, …, $x_{n}$ is $\overline{x}$, then mean of $x_{1} + k$, $x_{2} + k$, $x_{3} + k$, …, $x_{n} +k$ will be $\overline{x} +k$.
• If each value in the data gets multiplied or divided by a fixed value, then the mean also gets multiplied or divided by the same number. If the mean of $x_{1}$, $x_{2}$, $x_{3}$, …, $x_{n}$ is $\overline{x}$, then the mean of $k.x_{1}$, $k.x_{2}$, $k.x_{3}$, …, $k.x_{n}$ is $k.\overline{x}$. Similarly, the mean of $\frac{x_{1}}{k}$, $\frac{x_{2}}{k}$, $\frac{x_{3}}{k}$, …, $\frac{x_{n}}{k}$ will be $\frac{\overline{x}}{k}$.

## Arithmetic Mean Formula

The general formula for calculating the mean of the given data is $\overline{x} = \frac{\text{Sum of all observations}}{\text{Count of observations}}$.

## Arithmetic Mean for Ungrouped Data

If $x_{1}$, $x_{2}$, $x_{3}$, …, $x_{n}$ are $n$ data values then the arithmetic mean $\overline{x} = \frac{x_{1} + x_{2} + x_{3} + … + x_{n}}{n} = \frac{\sum {x_{i}}}{n}$.

### Examples

Ex 1: Find the arithmetic mean of $7$, $4$, $6$, $5$, $8$, $5$, $8$, $4$, $6$, $8$, $5$, $7$ and $5$.

Number of data items $n = 13$.

Sum of data item = $\sum {x_{i}} = 7 + 4 + 6 + 5 + 8 + 5 + 8 + 4 + 6 + 8 + 5 + 7 + 5 = 78$

Arithmetic Mean = $\frac {\sum {x_{i}}}{n} = \frac {78}{13} = 6$.

Ex 2: Find the arithmetic mean of $5.2$, $4.8$, $5.0$, $5.6$, $4.4$, $5.8$, $5.4$, and $4.2$.

Number of data items = $n = 8$.

Sum of data items = $\sum {x_{i}} = 5.2 + 4.8 + 5.0 + 5.6 + 4.4 + 5.8 + 5.4 + 4.2 = 40.4$

Arithmetic Mean = $\frac {\sum {x_{i}}}{n} = \frac {40.4}{8} = 5.05$.

Ex 3: The temperature (in degrees celsius) recorded for past $10$ days is $-4$, $-5$, $0$, $2$, $-1$, $3$, $0$, $-3$, $-2$, and $-1$. Find the mean temperature.

Number of days = $n = 10$.

Sum of data items = $\sum {x_{i}} = -4 + \left(-5 \right) + 0 + 2 + \left(-1 \right) + 3 + 0 + \left(-3 \right) + \left( -2 \right) + \left(-1 \right) = -11$.

Mean Temperature = $\frac {\sum {x_{i}}}{n} = \frac {-11}{10} = -1.1^{\circ}\text{C}$.

Ex 4: If the arithmetic mean of $12$, $19$, $18$, $17$, $12$, $14$, $x$, $16$, $18$, and $14$ is $15.5$, then find the value of $x$.

Arithmetic Mean = $15.5$

Number of data item = $n = 10$

Sum of data items = $\sum {x_{i}} = 12 + 19 + 18 + 17 + 12 + 14 + x + 16 + 18 + 14 = 140 + x$.

Arithmetic Mean = $\frac {\sum {x_{i}}}{n} = \frac{140 + x}{10}$

Therefore, $\frac {\sum {x_{i}}}{n} = \frac{140 + x}{10} = 15.5 => 140 + x = 15.5 \times 10 => 140 + x = 155 => x = 155 – 140 => x = 15$

## Arithmetic Mean for Grouped Data

There are three methods to find the arithmetic mean of grouped data.

• Direct method
• Short-Cut method
• Step-Deviation method

The choice of the method depends on the numerical values of data values and their frequencies(number of occurrences). If $x_{i}$ and $f_{i}$ are sufficiently small, the direct method is used. But, if they are numerically large, we use the assumed arithmetic mean method or step-deviation method.

### Direct Method

The frequency distribution table can be either discrete data or continuous data

#### Direct Method for Discrete Data

If $x_{1}$, $x_{2}$, $x_{3}$, …, $x_{n}$ are observations having respective frequencies $f_{1}$,  $f_{2}$, $f_{3}$, …, $f_{n}$, then the sum of observations will be $f_{1}x_{1} + f_{2}x_{2} + f_{3}x_{3} + … + f_{n}x_{n}$ and count of observations is equal to sum of all frequencies, i.e., $f_{1} + f_{2} + f_{3} + … + f_{n}$.

Therefore, arithmetic mean $\overline{x} = \frac{f_{1}x_{1} + f_{2}x_{2} + f_{3}x_{3} + … + f_{n}x_{n}}{f_{1} + f_{2} + f_{3} + … + f_{n}} = \frac{\sum{f_{i}x_{i}}}{\sum{f_{i}}}$.

The following steps are used to compute the arithmetic mean by the direct method.

Step 1: Prepare the frequency table in such a way that its first column consists of the observations $\left(x_{i} \right)$ and the second column the respective frequency $\left(f_{i} \right)$. Make one more column to include the product of $f_{i}$ and $x_{i}$, i.e., $f_{i}x_{i}$

Step 2: Add all the frequencies in the second column to obtain $\sum{f_{i}}$.

Step 3: Add all the $f_{i}x_{i}$ in the third column to obtain $\sum{f_{i}x_{i}}$.

Step 4: To calculate mean use the formula $\frac{\sum{f_{i}x_{i}}}{\sum{f_{i}}}$.

#### Examples

Ex 1: Find the mean height of students of the following.

Mean height of student $\overline{x} = \frac{\sum{f_{i}x_{i}}}{\sum{f_{i}}} = \frac {3509}{23} = 152.56 \text{cm}$

Ex 2: Find the arithmetic mean of the following distribution.

The arithmetic mean =$\frac{\sum{f_{i}x_{i}}}{\sum{f_{i}}} = \frac{1326}{50} = 26.52$

Ex 3: Find the value of $p$, if the mean of the distribution is $18$.

Therefore, $\frac{399 + 100p + 5p^{2}}{23 + 5p} = 18 => 399 + 100p + 5p^{2} = 18 \times (23 + 5p) => 399 + 100p + 5p^{2} = 414 + 90p$

$=> 399 + 100p + 5p^{2} – 414 – 90p = 0 => 5p^{2} + 10p – 15 = 0 => p^{2} + 2p – 3 = 0$

$=> p^{2} – p + 3p – 3 = 0 => p\left(p – 1 \right) + 3 \left(p – 1 \right) = 0 => (p – 1)(p + 3) = 0 => p = 1 \text{ or } p = -3$

Since, frequency cannot be negative, therefore, $p = 1$.

#### Direct Method for Continuous Data

The following steps are used to compute the arithmetic mean by the direct method.

Step 1: Prepare a table with four columns such that

• The first column contains Class Interval
• The second column contains frequencies $\left(f_{i} \right)$.
• The third column contains classmarks(or mid-values). The formula to compute the classmark is $\frac{\text{Lower Limit} + \text{Upper Limit}}{2}$.
• The fourth column contains the product of frequency and classmark, i.e., $f_{i}x_{i}$.

Step 2: Add all the frequencies in the second column to obtain $\sum{f_{i}}$.

Step 3: Add all the $f_{i}x_{i}$ in the third column to obtain $\sum{f_{i}x_{i}}$.

Step 4: To calculate mean use the formula $\frac{\sum{f_{i}x_{i}}}{\sum{f_{i}}}$.

#### Examples

Ex 1: The following table shows the distribution of weekly wages of workers in a factory.

Find the weekly wages.

Therefore, the mean weekly wages are $\overline{x} = \frac{\sum{f_{i}x_{i}}}{\sum{f_{i}}} = \frac{79900}{180} =$ ₹$443.89$. (Rounded off to the nearest paise).

Ex 2: The mean of the following frequency distribution is $62.8$ and the sum of all frequencies is $50$. Compute the missing frequencies $f_{1}$ and $f_{2}$.

Since, the sum of all frequencies is $50 => 5 + f_{1} + 10 + f_{2} + 7 + 8 = 50 =>f_{1} + f_{2} = 20$ ———- (1)

$\overline{x} = \frac{\sum{f_{i}x_{i}}}{\sum{f_{i}}} => 62.8 = \frac{2060 + 30f_{1} + 70f_{2}}{50} =>2060 + 30f_{1} + 70f_{2} = 62.8 \times 50$

$=>2060 + 30f_{1} + 70f_{2} = 3140 =>30f_{1} + 70f_{2} = 3140 – 2060 =>30f_{1} + 70f_{2} = 1080 =>3f_{1} + 7f_{2} = 108$  ———- (2)

Solving the equations (1) and (2), we get

$f_{1} = 8$ and $f_{2} = 12$.

Therefore, the missing frequencies are $8$ and $12$ respectively.

### Short-Cut method

In this method, an approximate mean ( called assumed mean ) is taken ( preferably near the middle ), say $\text{A}$ and we calculate the deviation $d_{i} = x_{i} – \text{A}$ for each value of $x_{i}$. Then the mean $\overline{x}$ is computed using the formula: $\overline{x} = \text{A} + \frac{\sum{f_{i}d_{i}}}{\sum{f_{i}}}$.

Note: The assumed mean $\text{A}$ may not be one of the observations $x_{i}$ i.e., $\text{A}$ may be any non-zero real number. The value of the mean does not depend upon the choice of $\text{A}$.

The following steps are used to compute the arithmetic mean by the shortcut method.

Step 1: Prepare the frequency table in such a way that its first column consists of the class interval, the second column the respective frequencies, and the third column for the classmarks (or mid-values).

Step 2: Choose the assumed mean $text{A}$ and take deviations $d_{i} = x_{i} – \text{A}$.  Write these deviations against the respective frequencies in the fourth column.

Step 3: The fifth column consists of the product of the frequency of each row with the respective deviation i.e., $f_{i}d_{i}$.

Step 4: Add all the frequencies in the second column to obtain $\sum{f_{i}}$.

Step 5: Add all the products in the fifth column to obtain $\sum{f_{i}d_{i}}$.

Step 6: To calculate mean, use the formula $\overline{x} = \text{A} + \frac{\sum{f_{i}d_{i}}}{\sum{f_{i}}}$.

### Examples

Ex 1: Let’s consider the above example to compute the mean using the assumed mean method.

The following table gives the distribution of weekly wages in rupees of workers in a certain factory.

Find the weekly wages.

First of all, we’ll calculate the classmark(or mid-value).

The classmarks are $325$, $375$, $425$, $475$, $525$ and $575$.

Let’s consider the assumed mean $A = 475$.

### What is the use of the arithmetic mean?

The arithmetic mean is a measure of central tendency. It allows us to know the center of the frequency distribution by considering all of the observations.

### What is the sum of deviations from the arithmetic mean?

The sum of deviations from the arithmetic mean is equal to zero.

## Conclusion

The arithmetic mean is the simplest and most widely used measure of a mean, or average. The arithmetic mean formula simply involves taking the sum of a group of numbers, then dividing that sum by the count of the numbers used in the series. It allows us to know the centre of the frequency distribution by considering all of the observations.